A student following the procedure in this experiment acquired temperature as a f
ID: 523668 • Letter: A
Question
A student following the procedure in this experiment acquired temperature as a function of time data in order to determine the heat capacity of the calorimeter. Based on a plot of these data and extrapolation back to time zero, the student determined the final temperature to be 14.3 degree C. The room temperature watere had volume of 49.2 ml . The cold water had volume of 49.7 ml . The student measured Th=22.4 degree C and Tc=2.9 degree C . Use these data to determine the heat capacity of the calorimeter in J/C degree
Explanation / Answer
Apply
Q = 0 since adiabatic
Qcold + Qhot + Qcal = 0
Qcold = mcold*C*(Tf-Tcold)
Qhot = mhot*C*(Tf-Thot)
Qcold = Ccal*(Tf-Tcal)
mcold*C*(Tf-Tcold) + mhot*C*(Tf-Thot) + Ccal*(Tf-Tcal) =0
substitute data
49.7*4.184*(14.3-2.9) + 49.2*(4.184)*(14.3-22.4) + Ccal*(14.3-25) = 0
49.2*(4.184)*(14.3-22.4) + Ccal*(14.3-25) = 49.7*4.184*(14.3-2.9)
solve for CcAl
Ccal = ( 49.7*4.184*(14.3-2.9) - 49.2*(4.184)*(14.3-22.4) )/(14.3-25)
Ccal = 377.3811 J/C
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